Toying with Voltage Ratings

Discussion in 'General Electronics Chat' started by xibalban, Apr 12, 2013.

  1. xibalban

    Thread Starter New Member

    Apr 4, 2013
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    I was wondering if the following makes any difference, technically.

    Say, some load is rated at 100 W at 230 Volts (resistive load assumed)
    By using Power formula:

    I=P/V [Power factor=1 assumed]
    So, I=100/230 = 435 mA

    Let's connect a 12 V source [battery] across this same load:

    I=P/V
    So, I=100/12 = 8.3 A

    Hence, if a 40 Ah, 12 V battery is connected across this load, it would last for approximately 5 hours:

    Total hours, h = Ah/A
    h = 40 Ah/8.3 A
    So, h = 4.8 hours

    I know that this setup would require thicker connecting wires due to higher amperage, but I have the following doubts:

    • Is it practical?
    • Can a 230 V rated load be connected to a 12 V source?
    • What happens if the load were to be inductive/capacitive?
    Please shoot your comments fellas!
     
  2. #12

    Expert

    Nov 30, 2010
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    You forgot to convert the resistive load to how many ohms it has. When you connect a 12 volt battery to something that can withstand 230 volts, a lot less current will flow.
     
  3. MrChips

    Moderator

    Oct 2, 2009
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    #12 is correct.

    Since power = V^2/R

    The new power at 12V is 100*(12/230)^2 = 0.272W

    instead of 100W.
     
  4. xibalban

    Thread Starter New Member

    Apr 4, 2013
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    Thanks, I realise that too now.

    Why aren't loads rated in impedence/resistance instead of common voltage and wattage ratings?
     
  5. #12

    Expert

    Nov 30, 2010
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    Because 99% of people don't know about Ohm's Law and Watt's Law, and sometimes, even the ones that do know make mistakes.:D

    Better to keep it as simple as 12V, 2A or 120VAC, 1200 watts, and sometimes, that still isn't simple enough for people that aren't into electronics at all.
     
  6. xibalban

    Thread Starter New Member

    Apr 4, 2013
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    Would my assumption about the Ah to h conversion have been valid, if we introduced a voltage converter after the battery i.e. 12 V to 230 V converter, and feed the load with rated voltage?
     
  7. #12

    Expert

    Nov 30, 2010
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    Yes, except for the inefficiency losses of the converter.
     
  8. gerty

    AAC Fanatic!

    Aug 30, 2007
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    That would change the dynamics of the whole load, and introduce the losses of the inverter.

    Ya beat me!!
     
  9. #12

    Expert

    Nov 30, 2010
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    "Some days the magic works and some days, it doesn't.":D

    (Chief Dan George in the movie, Little Big Man)
     
  10. gerty

    AAC Fanatic!

    Aug 30, 2007
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    I know it didn't take me 5 minutes to type that sentence:eek:
     
  11. xibalban

    Thread Starter New Member

    Apr 4, 2013
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    I see, so just in order to confirm:

    "A 40 Ah, 12 V battery would light a 230V, 100 W bulb for 4.8 hours provided the output voltage is converted from 12 V t0 230 V with 100% conversion efficiency"

    Right?
     
  12. #12

    Expert

    Nov 30, 2010
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    Right.......
     
  13. xibalban

    Thread Starter New Member

    Apr 4, 2013
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    Thank you, your help was/is highly appreciated. Just to add more info for the sake of other readers:

    "100 W load will draw 435 mA from a 230 V source (Inverter for instance) while at the same time would draw about 8.3 A from the 12 V battery (original source) or more if losses are taken into account"

    You guys made my day, thanks.
     
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